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Physics Question on Electromagnetic induction

A circular coil of radius 8 cm8\, cm, 400400 turns and resistance 2 Omega2 \,Omega is placed with its plane perpendicular to the horizontal component of the earths magnetic field. It is rotated about its vertical diameter through 180∘180^{\circ} in 0.30 s0.30\, s. Horizontal component of earth magnetic field at the place is 3×10−5 T3\times 10^{-5}\, T The magnitude of current induced in the coil is approximately

A

4×10−2 A4 \times 10^{-2}\,A

B

8×10−4 A8 \times 10^{-4}\,A

C

8×10−2 A8 \times 10^{-2}\,A

D

1.92×10−3 A1.92 \times 10^{-3}\,A

Answer

8×10−4 A8 \times 10^{-4}\,A

Explanation

Solution

Initial flux through the coil ϕi=BA cos θ\phi_{i}=BA\, cos\, \theta =3×10−5×π×(8×10−2)2×cos 0∘=3 \times 10^{-5}\times\pi\times(8\times10^{-2})^{2} \times cos\, 0^{\circ} =192 π×10−9 WB=192\, \pi \times 10^{-9}\,WB Final flux after the rotation ϕf=3×10−5×π×(8×10−2)2×cos 180∘\phi_{f}=3\times 10^{-5}\times\pi\times(8\times10^{-2})^{2}\times cos\, 180^{\circ} =−192 π×10−9 Wb=-192\, \pi \times 10^{-9}\,Wb ∴\therefore The magnitude of induced emf is ε=N∣dϕ∣dt=N∣ϕf−ϕi∣dt\varepsilon=N \frac{\left|d\phi\right|}{dt}=\frac{N \left|\phi_{f}-\phi_{i}\right|}{dt} =400×(384π×10−9)0.30=\frac{400\times\left(384\pi\times10^{-9}\right)}{0.30} =1.6×10−3 V=1.6\times 10^{-3}\,V Current, I=εRI=\frac{\varepsilon}{R} =1.6×10−32=\frac{1.6\times10^{-3} }{ 2} =8×10−4 A=8 \times10^{-4}\,A